$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a $1 \ L$ closed vessel and heated until the following equilibrium is attained:
${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$
If $50\%$ of $N_2O_{4(g)}$ dissociates at equilibrium,what will be the equilibrium constant (in $mol \ L^{-1}$)? (Mol. wt. of $N_2O_4 = 92$)

$(1) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \ ; \ K_1$
$(2) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \ ; \ K_2$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_2O_{(g)} \ ; \ K_3$
The equation for the equilibrium constant of the reaction
$2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$
$(K_4)$ in terms of $K_1$,$K_2$,and $K_3$ is

$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
In an equilibrium mixture,the partial pressures are
$P_{SO_{3}} = 43 \ kPa$,$P_{O_{2}} = 530 \ Pa = 0.53 \ kPa$,and
$P_{SO_{2}} = 45 \ kPa$. The equilibrium constant $K_{p} = ...... \times 10^{-2} \ kPa^{-1}$. (Nearest integer)

For a gaseous reversible reaction,the enthalpy of reaction at constant pressure is $1.8 \ Kcal/mol$ greater than that at constant volume at $300 \ K$. The value of $\left( \frac{K_P}{K_C} \right)$ for the reaction at $T = \left( \frac{1}{0.00821} \right) \ K$ is:

Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$